Show that $\sqrt{q}$ is irrational

Question

Let q be a positive integer such that $q \geq 2$ and such that for any integers a and b, if $q|ab$, then $q|a$ or $q|b$. Show that $\sqrt{q}$ is irrational.

Proof;

Let assume $\sqrt{q}$ is a rational number, where $n \neq 0$ and $\gcd (m,n)=1$, meaning $\sqrt{q} = \frac{m}{n} \Rightarrow q=\frac{m^2}{n^2} $

Since $n^2 \nmid m^2$, $q|m^2 \Rightarrow q|m$, so $m=qt$ where $t\in \mathbb{Z}$

By substitute $m=qt$ in the equation $qn^2 = m^2$, we get $n^2=qt^2$.

Since tells us that $q|n^2$ and $t^2|n^2$, it contradicts with the assumption $\gcd (m,n)=1$; therefore, $\sqrt{q}$ is irrational.

I get this proof with the assistant of the course, but is there any flaw or mistake? What are the other methods for proving this statement, can you at least give one different method? And how can I improve this proof?

@AJY Which satisfies the condition that if $q |ab$, then $q | a$ or $q |b$ — Jacob, Oct 26 '16 at 16:25
@Leth, see this. What we usually call prime positive integer is what in algebra is called irreducible element. It's a consequence of Fundamental theorem of arithmetic that prime and irreducible positive integers are one and the same thing. — Ennar, Oct 26 '16 at 16:26
@jacer21, but $4\mid 2\cdot 2$, and yet, $4\not\mid 2$. You are confusing existential and universal quantification. — Ennar, Oct 26 '16 at 16:27
@Ennar I basically understand nothing, Could you explain the case when q=4 ? — Our, Oct 26 '16 at 16:27
@Leth, read my comment above. Basically, $(\forall a,b)\ q\mid ab\implies q\mid a$ or $q\mid b$ is equivalent to $q$ being prime number. — Ennar, Oct 26 '16 at 16:29
@Leth, it doesn't matter if $q = 4$ satisfies given implication for some $a,b$. It doesn't satisfy it for all $a,b$. For example, $a=b=2$. — Ennar, Oct 26 '16 at 16:30
If $q$ is not a prime then for some $a,b>1$ we have $q=a\cdot b$, but $a<q$ and $b<q$ since they are factors of $q$. So $q$ divides neither. Take contrapositive of this and your condition basically says $q$ is a prime — PLE, Oct 26 '16 at 16:32
@Ennar, Firstly, the fact that q is prime explained with those theorems is something that I haven't studied, so using this knowledge doesn't help me.Secondly, so you are saying that the statement is wrong ? — Our, Oct 26 '16 at 16:35
@Leth, this is completely irrelevant, the statement is true. I was just addressing jacer21 as he tried to make counterexample to the claim by using that $\sqrt 4$ is not irrational. This counterexample doesn't work since $q$ must be prime by explicit assumption, and $4$ is not prime. I've given you links since you asked me about it, but as I said, it is completely irrelevant for the proof you gave, which looks ok. — Ennar, Oct 26 '16 at 16:38
The tag ([tag:logic]) should be used for questions about mathematical logic, see the tag-info. — Martin Sleziak, Oct 27 '16 at 06:06

score 2 · Answer 1 · answered Oct 26 '16 at 16:38

2

Proof using Bézout's Identity

For $\sqrt{q}$ to be irrational, $q$ must not be a perfect square. Thus, we only concern us with non-perfect square $q$.

Assume that $\sqrt{q}$ is rational. Therefore $\sqrt{q} = \frac{m}{n}$ where $\gcd{(m, n)} = 1$.

By Bézout's Identity, there exist integers $x$ and $y$ such that $mx + ny = 1$

Now $\sqrt{q} = \sqrt{q}(1) = \sqrt{q}(mx + ny) = (\sqrt{q}m)x + (\sqrt{q}n)y = qnx + my = \text{an integer}$

Which leads us to a contradiction, since we initially assumed that $q$ wasn't a perfect square. Hence, $\sqrt{q}$ is irrational.

answered Oct 26 '16 at 16:38

Ojas

147

I did not understand the part $(\sqrt{q}m)x + (\sqrt{q}n)y = qnx + my$ – Our Oct 26 '16 at 16:41
1

Notice that from our initial assumption we get $m = \sqrt{q} n$ – Ojas Oct 26 '16 at 16:43
1

@Leth See also this answer, which explains how to view such Bezout-based proofs as descent on denominators. See also many other proofs in that thread. – Bill Dubuque Oct 26 '16 at 20:18

Jack D'Aurizio · Answer 2 · 2016-10-27T05:09:25.450

1

We just have to show that for any prime number $p$, $\sqrt{p}\not\in\mathbb{Q}$.
If we assume $\sqrt{p}=\frac{a}{b}$ with $a,b\in\mathbb{Z}^+$ we get the identity $pb^2=a^2$.
For any $n\in\mathbb{Z}^+$, let $\nu_p(n)=\max\{m\in\mathbb{N}: p^m\mid n\}$. The identity $pb^2=a^2$ implies $$ \nu_p(pb^2) = \nu_p(a^2) $$ but that is impossible, since $\nu_p(a^2)$ is an even number and $\nu_p(pb^2)=1+\nu_p(b^2)$ is an odd number. $\sqrt{p}\not\in\mathbb{Q}$ trivially follows.

edited Oct 27 '16 at 05:09

answered Oct 26 '16 at 22:12

Jack D'Aurizio

353,855

why just the odd numbers ? – Our Oct 27 '16 at 04:40
@Leth: I misread "for any $q\color{red}{>}2$ such that $\ldots$" in the main question, now fixed. It is not really relevant, indeed. – Jack D'Aurizio Oct 27 '16 at 05:10
but I still don't understand why the p is prime.Above Ennar gave some info about it, but it is about something that I haven't studied. – Our Oct 27 '16 at 09:18
@Leth: $$q\mid(ab)\implies q\mid a\text{ xor } q\mid b$$ is the very definition of prime. As an alternative, it is easy to show that only prime numbers (by your current definition) may have such property. – Jack D'Aurizio Oct 27 '16 at 09:20
Can you prove it ? or give a reference ? – Our Oct 27 '16 at 11:54
1

@Leth: https://en.wikipedia.org/wiki/Prime_element – Jack D'Aurizio Oct 27 '16 at 15:18

Show that $\sqrt{q}$ is irrational

2 Answers2